Is this operator a distribution?

Question

Is this operator: $$T: \mathcal{C}^{\infty}_0 \ni \varphi \to \lim_{x \to \infty} x^2 e^{-x} \varphi'(x) \in \mathbb{R}$$

a distribution (generalized function)?

I need to check two things: whether it is linear and continuous.

Continuous in the sense that for $\varphi_i \in \mathcal{C}^{\infty}_0$ such that $\forall i: supp \varphi_i \subset [-N, N] $ and $max_{x \in [-N, N]} |\varphi_i^{(\alpha)}(x) - \varphi^{(\alpha)}(x) | \to 0 \ (i \to \infty)$ we have $T(\varphi_i) \to T(\varphi) \ (i \to \infty)$.

I can see that this operator is linear, but I have trouble with proving the continuity.

We have that $$T(\varphi_i) = \lim_{x \to \infty} x^2 e^{-x} \varphi_i'(x)$$

Does it tend to $\lim_{x \to \infty} x^2 e^{-x} \varphi_i'(x)$ as $i \to \infty$?

$$|T(\varphi_i) - T(\varphi)| = |\lim_{x \to \infty} x^2 e^{-x} \varphi_i'(x) - \lim_{x \to \infty} x^2 e^{-x} \varphi'(x)| = |\lim_{x \to \infty} x^2 e^{-x} (\varphi_i'(x) - \varphi'(x))|$$

How can I use this: $max_{x \in [-N, N]} |\varphi_i^{(\alpha)}(x) - \varphi^{(\alpha)}(x) | \to 0 \ (i \to \infty)$ so that everything stays true and correct?

EDIT:

Following your comments, I've calculated:

$$T(\varphi) = \lim_{x \to \infty} \frac{x^2}{e^x} \varphi'(x)$$

Now, $\varphi \in \mathcal{C}_0^{\infty}$, so $\varphi ' \in \mathcal{C}_0^{\infty}$, so $\lim_{x \to \infty} \varphi '(x)=0$ and so does $\frac{x^2}{e^x}$.

So $T(\varphi) = 0$.

Is that correct?

Didn't I miss anything?

Answer 1

Your functional assigns to every test function the value $0$. Since $$ \mathrm{e}^{-x}x^2\varphi(x)=0, $$ if $x$ is sufficiently large.